Chapter 10.1, Problem 48E

To determine

To find: Total distance travelled by ball.

Answer to Problem 48E

Total distance travelled by ball is $D=16$ m.

Explanation of Solution

Given:

Height from which ball is dropped is 4 m.

Ball rebounds to a height 0.06 h after striking the floor.

Formula used:

A geometric progression with initial term $a$ and common ratio $r$ Then the sum of this geometric progression with $r<1$ is,

  $S=\frac{a}{1-r}$

Calculation:

Distance travelled by ball before 1^st striking is 4 m.

Now,

Height attained after 1^st strike is $0.6\times 4=2.4\text{m}$

So, distance travelled by ball after 1^st and before 2^nd strike is

  $\left(0.6\times 4\right)+\left(0.6\times 4\right)=2\left(0.6\times 4\right)\text{m}$

Now,

Height attained after 2^nd strike is $\left(0.6\times 4\right)\times 0.6={0.6}^2\times 4$

Distance travelled by ball after 2^ndand before 3^rd strike is

  $\left({0.6}^2\times 4\right)+\left({0.6}^2\times 4\right)=2\left({0.6}^2\times 4\right)\text{m}$

Similarly,

Distance travelled by ball after 3^rdand before 4^thstrike is

  $2\left({0.6}^3\times 4\right)\text{m}$

And it continues.

Hence, total distance travelled by the ball is

  $D=4+\left[2\left(0.6\right)4+2{\left(0.6\right)}^{2}4+2{\left(0.6\right)}^{3}4+\mathrm{...}\right]$

Taking,

  $S=2\left(0.6\right)4+2{\left(0.6\right)}^{2}4+2{\left(0.6\right)}^{3}4+\mathrm{...}$

Here, it is a geometric progression with initial term $a=4.8$ and common ratio $r=0.6$

The sum of this geometric progression with $r<1$ is,

  $\begin{array}{l}S=\frac{a}{1-r}\\ =\frac{4.8}{\left(1-0.6\right)}\\ =12\end{array}$

Hence, total distance travelled is

  $\begin{array}{l}D=4+S\\ =4+12\\ \Rightarrow D=16\end{array}$

Calculation: Total distance travelled by ball is $D=16$ m.